To calculate the [tex]$\Delta H$[/tex] for the given reaction:
[tex]\[
CS_2 + 2 H_2O \rightarrow CO_2 + 2 H_2S
\][/tex]
we will use the two given reactions and manipulate them to match the desired equation. The reactions provided are:
[tex]\[
1. \quad H_2S + 1.5 O_2 \rightarrow H_2O + SO_2 \quad \Delta H = -563 \, \text{kJ}
\][/tex]
[tex]\[
2. \quad CS_2 + 3 O_2 \rightarrow CO_2 + 2 SO_2 \quad \Delta H = -1075 \, \text{kJ}
\][/tex]
### Step-by-Step Solution:
1. Reverse the first given reaction:
Reverse the reaction:
[tex]\[
H_2O + SO_2 \rightarrow H_2S + 1.5 O_2
\][/tex]
When you reverse a reaction, the sign of [tex]$\Delta H$[/tex] changes:
[tex]\[
\Delta H = +563 \, \text{kJ}
\][/tex]
2. Scale the reversed reaction:
Our target reaction has 2 moles of [tex]$H_2S$[/tex] and [tex]$H_2O$[/tex]. To match this, we need to multiply the reversed reaction by 2:
[tex]\[
2 (H_2O + SO_2 \rightarrow H_2S + 1.5 O_2)
\][/tex]
This gives us:
[tex]\[
2 H_2O + 2 SO_2 \rightarrow 2 H_2S + 3 O_2
\][/tex]
The enthalpy change for this scaled reaction will be:
[tex]\[
\Delta H = 2 \times 563 \, \text{kJ} = 1126 \, \text{kJ}
\][/tex]
3. Combine with the second given reaction:
Now, we can combine this scaled reaction with the second given reaction. The combined reactions are:
[tex]\[
CS_2 + 3 O_2 \rightarrow CO_2 + 2 SO_2 \quad \Delta H = -1075 \, \text{kJ}
\][/tex]
[tex]\[
2 H_2O + 2 SO_2 \rightarrow 2 H_2S + 3 O_2 \quad \Delta H = 1126 \, \text{kJ}
\][/tex]
4. Sum the reactions:
When we sum these reactions, the [tex]$O_2$[/tex] and [tex]$SO_2$[/tex] terms will cancel out, and we get:
[tex]\[
CS_2 + 2 H_2O \rightarrow CO_2 + 2 H_2S
\][/tex]
The total enthalpy change ([tex]$\Delta H$[/tex]) for the overall reaction is the sum of the enthalpy changes of the individual reactions:
[tex]\[
\Delta H = -1075 \, \text{kJ} + 1126 \, \text{kJ} = 51 \, \text{kJ}
\][/tex]
### Final Answer:
[tex]\[
\Delta H \, \text{for} \, CS_2 + 2 H_2O \rightarrow CO_2 + 2 H_2S \, \text{is} \, +51 \, \text{kJ}
\][/tex]
